MP 6..1So, do we lose the capability to do work when we have an
irreversible process and entropy increases?

Absolutely. We will see this in a more general fashion very soon.
The idea of lost work is one way to view what ``entropy is all
about.''

MP 6..2Why do we study cycles starting with the Carnot cycle? Is it
because it is easier to work with?

Carnot cycles are the best we can do in terms of efficiency. We use
the Carnot cycle as a standard against which all other cycles are
compared. We will see in class that we can break down a general
cycle into many small Carnot cycles. Doing this we can gain insight
in which direction the design of efficient cycles should go.

I find
-
diagrams useful, especially in dealing with
propulsion systems, because the difference in stagnation
can be
related (from the Steady Flow Energy Equation) to shaft work and
heat input. For processes that just have shaft work (compressors or
turbines) the change in stagnation enthalpy is the shaft work. For
processes that just have heat addition or rejection at constant
pressure, the change in stagnation enthalpy is the heat addition or
rejection.

I think working with these diagrams may be the most useful way to
achieve this objective. In doing this, the utility of using these
coordinates (or
-
coordinates) should also become clearer. I
find that I am more comfortable with
-
or
-
diagrams
than with
-
diagrams, especially the latter because it conveys
several aspects of interest to propulsion engineers: work produced
or absorbed, heat produced or absorbed, and loss.

Let's pick an example process involving a chamber filled with a
compressible gas and a piston. We assume that the chamber is
insulated (so no heat transfer to or from the chamber) and the
process is thus adiabatic. Let us also assume that the piston is
ideal, such that there is no friction between the walls of the
chamber and the piston. The gas is at some temperature
. We now
push the piston in and compress the gas. The internal energy of the
system will then increase by the amount of work we put in and the
gas will heat up and be at higher pressure. If we now let the piston
expand again, it will return to its original position (no friction,
ideal piston) and the work we took from the environment will be
returned (we get the exact same amount of work back and leave no
mark on the environment). Also, the temperature and the pressure of
the gas return to the initial values. We thus have an adiabatic
reversible process. For both compression and expansion we have no
change in entropy of the system because there is no heat transfer
and also no irreversibility. If we now draw this process in the
-
or
-
diagram we get a vertical line since the entropy
stays constant:
or
and we can
also call this process an isentropic process.

MP 6..7If there is an ideal efficiency for all cycles, is there a
maximum work or maximum power for all cycles?

Yes. As with the Brayton cycle example, we could find the maximum as
a function of the appropriate design parameters.

MP 6..8Do we ever see an absolute variable for entropy? So far, we
have worked with deltas only.

It is probably too strong a statement to say that for ``us'' the
changes in entropy are what matters, but this has been my experience
for the type of problems aerospace engineers work on. Some values of
absolute entropy are given in Table A.8 in SB&VW. We will also see,
in the lectures on Rankine cycles, that the entropy of liquid water
at a temperature of
and a pressure of
has been specified as zero for problems
involving two-phase (steam and water) behavior.

Both of these are true and both can always be used. The first is the
definition of entropy. The second is a statement of how the entropy
behaves. Section 6.5 attempts to make the
relationship clearer through the development of the equality
.

MP 6..10For irreversible processes, how can we calculate
if not
equal to
?

We need to define a reversible process between the two states in
order to calculate the entropy (see
MP 6.9). See VN Chapter 5 (especially) for
discussion of entropy or Section 6.5. If you are
still in difficulty, come and see me.

No. Entropy is a function of state (see Gibbs) and thus
is path independent. For example we could have three different paths
connecting the same two states and therefore have the same change in
entropy

MP 6..12Are
and
the
and
going from the final state back to the initial state?

Yes. We have an irreversible process from state 1 to state 2. We
then used a reversible process to restore the initial state 1 (we
had to do work on the system and extract heat from the system).

MP 6..13Why do we find stagnation enthalpy if the velocity never
equals zero in the flow?

The stagnation enthalpy (or stagnation temperature) is a useful
reference quantity. Unlike the static temperature it does not vary
along a streamline in an adiabatic flow, even if irreversible. It
was thus the natural reference temperature to use in describing the
throttling process. In addition, changes in stagnation enthalpy (or
stagnation temperature) are direct representations of the shaft work
or heat associated with a fluid component. The static enthalpy is
not, unless we assume that changes in
are small. Measurement of
stagnation temperature thus allows direct assessment of shaft work
in a turbine or compressor, for example.

Because for a steady adiabatic flow with no shaft work done the
Steady Flow Energy Equation yields constant stagnation enthalpy even
though the flow processes might not be reversible (see notes). For a
perfect gas with constant specific heats,
, thus the
stagnation temperature remains constant.

MP 6..15How do you tell the difference between shaft work/power and
flow work in a turbine, both conceptually and mathematically?

Let us look at the expansion of a flow through a turbine using both
the control mass approach and the control volume approach. Using the
control mass approach we can model the situation by tracking
of air as follows: state 1 -- before the expansion
we have
of air upstream of the turbine. We then push
the gas into the turbine and expand it through the blade rows. After
the expansion we take
of air out of the turbine and
the mass of air is downstream of the turbine -- state 2. The work
done by the gas is work done by the turbine (blades moved around by
the gas) plus the work done by pressures (flow work). Using
the first law we can then write for the change of internal energy of
of air:

When entering the turbine, the fluid has to push the
surroundings out of the way to make room for itself (it has volume
and is at
) -- the work to do this is
. When
leaving the turbine the fluid is giving up room and the work to keep
that volume
at pressure
is freed; thus
. We can
then write for the shaft work

The right hand side of the above equation is the change in enthalpy
. This is another example to show how useful enthalpy
is (enthalpy is the total energy of a fluid: the internal energy
plus the extra energy that it is credited with by having a volume
at pressure
). The shaft work output by the turbine is equal
to the change in enthalpy (enthalpy contains the flow work!).

We can also solve this problem by using the 1st law
in general form (control volume approach):

Note that in this form the flow work is buried in
already! For
this turbine, we can drop the unsteady term on the left and neglect
heat fluxes (adiabatic turbine), shear work and piston work (no
pistons but blades, so we keep the shaft work). Further we assume
that changes in potential energy and kinetic energy are negligible
and we obtain for
air

We obtain the
same result as before:
.

MP 6..16Is the entropy change in the equations two lines above the
total entropy change? If so, why does it say
?

The entropy change in question is the entropy change due to the heat
produced by friction only.

When we wrote this equality, we were considering a system that was
returned to its original state, so that there were no changes in any
of the system properties. All evidence of irreversibility thus
resides in the surroundings.

MP 6..18In discussing the terms ``closed system'' and ``isolated
system,'' can you assume that you are discussing a cycle or not?

The terms closed system and isolated system have no connection to
whether we are discussing a cycle or not. They are attributes of a
system (any system), whether undergoing cyclic behavior, one-way
transitions, or just sitting there.

Entropy is a state function. If the process is cyclic, initial and
final states are the same. So, for a cyclic process,
.

MP 6..20In a real heat engine, with friction and losses, why is
still if
?

The change in entropy during a real cycle is zero because we are
considering a complete cycle (returning to initial state) and
entropy is a function of state (holds for both ideal and real
cycles!). Thus if we integrate
around the real
cycle we will obtain
. What essentially
happens is that all irreversibilities (
's) are turned into
additional heat that is rejected to the environment. The amount of
heat rejected in the real ideal cycle
is going
to be larger than the amount of heat rejected in an ideal cycle
:

We will see this better using the
-
diagram. The change of
entropy of the surroundings (heat reservoirs) is
. So
even for real cycles, but
.

MP 6..21Is it safe to say that entropy is the
tendency for a system to go into disorder?

Entropy can be given this interpretation from a statistical
perspective, and this provides a different and insightful view of
this property. At the level in which we have engaged the concept,
however, we focus on the macroscopic properties of systems, and
there is no need to address the idea of order and disorder; as we
will see, entropy is connected to the loss of our ability to do
work, and that is sufficient to make it a concept of great utility
for the evaluation and design of engineering systems. We will look
at this in a later lecture. If you are interested in pursuing this,
places to start might be Great Ideas in Physics by Lightman
(paperback book by an MIT professor), or Modern
Thermodynamics by Kondepudi and Prigogine.

MP 6..22Isn't it possible for the mixing of two gases to go from the
final state to the initial state? If you have two gases in a box,
they should eventually separate by density, right?

Let us assume that gas
is oxygen and gas
is nitrogen. When
the membrane breaks the entire volume will be filled with a mixture
of oxygen and nitrogen. This may be considered as a special case of
an unrestrained expansion, for each gas undergoes an unrestrained
expansion as it fills the entire volume. It is impossible for these
two uniformly mixed gases to separate without help from the
surroundings or environment. A certain amount of work is necessary
to separate the gases and to bring them back into the left and right
chambers.

MP 6..23How can
be the maximum turbine inlet temperature?

I agree that the
is a temperature ratio. If we assume
constant ambient temperature then this ratio reflects the maximum
cycle temperature. The main point was to emphasize that the higher
your turbine inlet temperature the higher your power and efficiency
levels.

MP 6..24When there are losses in the turbine that shift the
expansion in
-
diagram to the right, does this mean there is
more work than ideal since the area is greater?

We have to be careful when looking at the area enclosed by a cycle
or underneath a path in the
-
diagram. Only for a
reversible cycle, the area enclosed is the work done by the
cycle (see notes Section 6.3). Looking at the
Brayton cycle with losses in compressor and turbine the net work is
the difference between the heat absorbed and the heat rejected (from
the 1st law). The heat absorbed can be found by
integrating
along the heat addition process. The heat
rejected during the cycle with losses in compressor and turbine is
larger than in the ideal cycle (look at the area underneath the path
where heat is rejected, this area is larger than when there are no
losses (when
). See also
muddy point 6.1. So we get less net
work if irreversibilities are present. It is sometimes easier to
look at work and heat (especially shaft work for turbines and
compressors) in the
-
diagram because the enthalpy difference
between two states directly reflects the shaft work (remember,
enthalpy includes the flow work!) and / or heat transfer.

MP 6..25For an afterburning engine, why must the nozzle throat area
increase if the temperature of the fluid is increased?

The Mach number of the flow is unity at the throat with and without
the afterburner lit. The ratio of static pressure to stagnation
pressure at the throat is thus the same with and without the
afterburner lit. The ratio of static temperature to stagnation
temperature at the throat is thus the same with and without the
afterburner lit.

The flow through the throat is

The flow through the throat thus scales as

From what we have said, however, the pressure at the throat is the
same in both cases. Also, we wish to have the mass flow the same in
both cases in order to have the engine operate at near design
conditions. Putting these all together, plus use of the idea that
the ratio of stagnation to static temperature at the throat is the
same for both cases gives the relation

The necessary area to pass the flow is proportional to the square
root of the stagnation temperature. If too much fuel is put into the
afterburner, the increase in area cannot be met and the flow will
decrease. This can stall the engine, a serious consequence for a
single engine fighter.

MP 6..26Why doesn't the pressure in the afterburner go up if heat is
added?

From discussions after lecture, the main point here seems to be that
the process of heat addition in the afterburner, or the combustor,
is not the same as heat addition to a gas in a box. In that case the
density (mass/volume) would be constant and, from
,
increasing the temperature would increase the pressure. In a
combustor, the geometry is such that the pressure is approximately
constant; this happens because the fluid has the freedom to expand
so the density decreases. From the equation
if the
temperature goes up, the density must go down.

MP 6..27Why is the flow in the nozzle choked?

As seen in Unified, choking occurs when the stagnation to static
pressure ratio
gets to a certain value,
for gas
with
of 1.4. Almost all jet aircraft aircraft operate at
flight conditions such that this is achieved. If you are not
comfortable with the way in which the concepts of choking are laid
out in the Unified notes, please see me and I can give some
references.

MP 6..28What's the point of having a throat if it creates a
retarding force?

As shown in Unified, to accelerate the flow from subsonic to
supersonic, i.e., to create the high velocities associated with high
thrust, one must have a converging-diverging nozzle, and
hence a throat.